$x^3 + 2x^2 + 5x + 2\cos x = 0$ 
$x^3 + 2x^2 + 5x + 2\cos x = 0$

How do I find the number of solutions of  this equation (in $[0, 2\pi]$) without a graph?
Attempt: 
The equation simplifies to $x(x^2 + 2x + 5)=- 2\cos x $
Minima of the quadratic occurs at $x= -1$ and it's value is $4$
Minima of $-2\cos x$ is $-2$
 A: 
Let $$f(x)=x^3 + 2x^2 + 5x + 2\cos x$$ 

Then $$f'(x)=3x^2+4x-2\sin x+5$$  Since $$3x^2+4x+5>2\ge 2\sin x$$ for all $x\in\mathbb{R}$, we have that $$f'(x)>0$$ for all $x\in\mathbb{R}$. Thus $f$ is monotonically increasing for all real $x$.
Now as $$f(-1)=-1+2-5+1.08...=-2.91...<0,$$ $$f(0)=0+0+0+2=2>0$$ and $f$ is continuous for all values of $x$, by the Intermediate Value Theorem the equation $$f(x)=0$$ has only one solution.
P.S. If you are interested, the value of $x$ such that $f$ crosses the $x$-axis is around $-0.421$: https://www.desmos.com/calculator/2vnhsxqgz2
A: Look first at the cubic part. You see that it is always increasing, and the minimum derivative is found at the inflection point, which is at $x=-2/3$. The derivative there has value $11/3>2$,  so that the whole function, including the trigonometric part, is increasing. Thus it has only one root.
A: You will have to sketch a graph to find the number of solutions.
You don't need to get it exactly detailed and in the right scale, so you can do it on rough paper pretty easily without a calculator.
But how?
You identified the first part correctly, separating the two different functions from each other:
$$x^3 + 2x^2 + 5x=- 2\cos x $$
Now, sketching the RHS on rough paper is very easy. We'll focus on the polynomial in the LHS.
Turning point of the cubic occurs at $3x^2+4x+5=0$ which has zero solutions ($\because D<0$), hence, there is no turning point i.e. the function $f(x)$ is monotonic. The coefficient of $x^3$ is $>1$, which implies the cubic starts from $-\infty$ when $x\rightarrow-\infty$, crosses the y-axis at $(0,5)$ and goes till $+\infty$ when $x\rightarrow+\infty$. 
Now, $f(-1)=-1+2-5=-4<-2$ and $f(x)$ $\forall x<-1$ is less than $-2$. Hence, notice $f(x)$ has intersected $y=-\cos x$ in interval (0,1) .
Knowing the nature of LHS and RHS, we can easily comment that there will not be any more points of intersection.
Problem solved.
A: $$y_1=-x^3-5x$$ and $$y_2=2x^2+2\cos x$$ 
for $x\ge0,  y_1<0, y_2>0$
for $-1<x<0$ both function decrease and have only one intersection
for $x\le-1$ there are no intersection $$-x^3-5x<2x^2+2\cos x$$
